Four coins with reflip problem?

I came across the following problem today.

Flip four coins. For every head, you get $$\1$$. You may reflip one coin after the four flips. Calculate the expected returns.

I know that the expected value without the extra flip is $$\2$$. However, I am unsure of how to condition on the extra flips. I am tempted to claim that having the reflip simply adds $$\\frac{1}{2}$$ to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong. I am also told the correct returns is $$\\frac{79}{32}$$ and I have no idea where this comes from.

• "I am tempted to claim that having the reflip simply adds $1/2$ to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong" - did you do the maths for this? It sounds like the right logic and would have given you the right answer so if it didn't work out you might want to share your calculations so we can tell you where it went wrong... Sep 19, 2018 at 10:54
• The question as stated doesn't require that the reflip be a tail... Of course, most people would not reflip a head for obvious reasons, but strictly speaking that is not required in the problem statement. With that literal reading, I don't think the expected value would change in light of a reflip... Sep 19, 2018 at 12:22
• Another way of thinking about this: you can flip four coins and re-flip if at least one is tails, or you can flip 5 coins and count the right-most coin only when one of the four left-most coins is tails. This lets you break the problem into 32 possibilities. Sep 19, 2018 at 13:34

Your temptation is right and your gut is wrong. You do get an extra $\frac12$ if you got tails at least once. The probability that you don't have a tail to reflip is $\frac1{16}$, so you get an extra $\frac12\left(1-\frac1{16}\right)=\frac{15}{32}$. This added to the base expectation of $2 = \frac{64}{32}$ gives $\frac{79}{32}$.
• Could you explain why this approach is wrong? $p(0h) = \binom{4}{0}0.5^4*1\$$\rightarrow p(1h) = \binom{4}{1}0.5^4*2\$$$\rightarrowp(2h) = \binom{4}{2}0.5^4*3\$$\rightarrow p(3h) = \binom{4}{3}0.5^4*4\$$ $\rightarrow$ $p(4h) = \binom{4}{4}0.5^4*4\$$\rightarrow (of course 1 is added for every case in which we have at least 1 tail that could be flipped) Sep 28, 2021 at 11:12 Expectation of first 4 flips is 2. Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5. Expectation of the fifth flip, condition on there is no tail, is 0. Probability there is a tail in the first 4 flips is 15/16. Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32 Let X_i=1 if i-th toss is head and 0 otherwise. The reward is$$\sum_{i=1}^4X_i + X_5\left( 1-\prod_{i=1}^4X_i\right)=\sum_{i=1}^5X_i-\prod_{i=1}^5X_i$$Hence$$\mathbb{E}\left(\sum_{i=1}^5X_i-\prod_{i=1}^5X_i\right)=\left(\sum_{i=1}^5\mathbb{E}[X_i]-\prod_{i=1}^5\mathbb{E}[X_i]\right)=\frac52-\frac1{32}=\frac{79}{32}$$• Very clear, thank you! Sep 19, 2018 at 8:42 Your answer is$\frac {80}{32}$. The$\frac1 {32}$is from where all four flips are heads. Your gut is wrong, as pointed out already. The possible outcomes from the initial flip are: "4 heads" x 1 "3 heads" x 4 "2 heads" x 6 "1 heads" x 4 "0 heads" x 1 This gives an expected return of (4 + 12 + 12 + 4 + 0)/16 = 2 If you add 0.5 to each case except 4 heads, you get (4 + 14 + 15 + 6 + 0.5)/16 = 79/32 This particular problem can be solved simply by flipping five coins but winning a maximum of \$4.
That gives an expected value of: $$\frac{1\cdot 0 + 5\cdot 1 + 10\cdot 2 + 10\cdot 3 + (5+1)\cdot 4}{2^5}=\frac{79}{32}.$$